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Phthalate buffer, Manganese Dioxide and Hydrogen peroxide = ? Rate Topic: -----

#1 ssrnightingale 


Lepton
So, as an A level chemistry student, I'm conducting an experiment into the decomposition of hydrogen peroxide, using manganese dioxide. I'm currently investigating the pH, and I have some very confusing results which I don't actually have any explanation for as of yet. Hydrogen peroxide is supposedly more stable at lower pHs, yet at pH 4, (with a phthalate buffer) more gas is being produced and at a quicker rate than at pH 6.

Can anybody help me and help me explain this?
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#2 foodchain 


Organism

ssrnightingale said:

So, as an A level chemistry student, I'm conducting an experiment into the decomposition of hydrogen peroxide, using manganese dioxide. I'm currently investigating the pH, and I have some very confusing results which I don't actually have any explanation for as of yet. Hydrogen peroxide is supposedly more stable at lower pHs, yet at pH 4, (with a phthalate buffer) more gas is being produced and at a quicker rate than at pH 6.

Can anybody help me and help me explain this?


Have you tried setting up a couple different sets which differ on just one variable? I mean this is probably a simple question really for some on here, but maybe you could see if there is a difference on how you are combining it in the solution, as are you pouring it in while stirring to just simply dripping it in. Part of reactions as you probably know also depend on how energetic the collisions are is all.

It might help if you could write out the balanced equation you are going for.
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#3 ssrnightingale 


Lepton
Equation:
2 H2O2 -> 2 H2O + O2

I'm using manganese dioxide as my catalyst, and I'm injecting my Hydrogen Peroxide.

I think I'll try that tomorrow, but due to time constraints I can't really do much investigation into the matter.
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#4 foodchain 


Organism

ssrnightingale said:

Equation:
2 H2O2 -> 2 H2O + O2

I'm using manganese dioxide as my catalyst, and I'm injecting my Hydrogen Peroxide.

I think I'll try that tomorrow, but due to time constraints I can't really do much investigation into the matter.


I was wondering on the catalyst part but did not bother to ask. That will have a huge impact as in some cases it can instantly change things big time in a reaction.

Yet I thought you were having a buffer to aid the ph in solution, and then adding one to study a decomposition reaction, is it producing heat? Like in the above equation, in the book do you also get heat from the reaction?

O2 will not exist as a liquid unless things are rather cold I would think, so all that liberated matter will want to escape.

You might want to do the moles and all of that in relation to your solution amount to see if you can estimate how much O2 is being produced.

I also do not remember if O2 is polar or non and how that would matter in relation to all of the other stuff.
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#5 ssrnightingale 


Lepton
Yes, I'm using the phthalate buffer to get the ph to 4. The thing is, my research tells me that hydrogen peroxide is more stable in more acidic conditions- hence why I'm so confused.

Wouldn't O2 be one of those with an instantaneous dipole?

I'm conducting my experiment in a 20C water bath, so I'm trying to control the temperature, and though it is an exothermic reaction, very little heat is actually being released.
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#6 foodchain 


Organism

ssrnightingale said:

Yes, I'm using the phthalate buffer to get the ph to 4. The thing is, my research tells me that hydrogen peroxide is more stable in more acidic conditions- hence why I'm so confused.

Wouldn't O2 be one of those with an instantaneous dipole?

I'm conducting my experiment in a 20C water bath, so I'm trying to control the temperature, and though it is an exothermic reaction, very little heat is actually being released.


Right, I think I understand, but the h202 is being broken down at a rate I would think. As far as temperature is concerned I think it has to be really cold before you can get liquid oxygen, and O2 is the state I think for oxygen to be in naturally. Plus in solution, depending on how much, you can only have so much of something dissolved in a solution at a giving time, so you might have reached a saturation point already, that's another reason to look at concentrations and what not from a perspective of moles and ratios, yes fractions and lots of them:D

I don't quite get the dipole question, plus I forgot most the math to figuring out polarity and molecular stuff like structure.
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#7 hermanntrude 


Icon
Chemistry Expert
The decomposition of H2O2 can be catalysed by many things. Perhaps one component of your buffer solution is acting as a catalyst?
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